Problem: Boboniu gives you

  * r red balls, 
  * g green balls, 
  * b blue balls, 
  * w white balls. 



He allows you to do the following operation as many times as you want: 

  * Pick a red ball, a green ball, and a blue ball and then change their color to white. 



You should answer if it's possible to arrange all the balls into a palindrome after several (possibly zero) number of described operations. 

Input

The first line contains one integer T (1≤ T≤ 100) denoting the number of test cases.

For each of the next T cases, the first line contains four integers r, g, b and w (0≤ r,g,b,w≤ 10^9).

Output

For each test case, print "Yes" if it's possible to arrange all the balls into a palindrome after doing several (possibly zero) number of described operations. Otherwise, print "No".

Example

Input


4
0 1 1 1
8 1 9 3
0 0 0 0
1000000000 1000000000 1000000000 1000000000


Output


No
Yes
Yes
Yes

Note

In the first test case, you're not able to do any operation and you can never arrange three balls of distinct colors into a palindrome.

In the second test case, after doing one operation, changing (8,1,9,3) to (7,0,8,6), one of those possible palindromes may be "rrrwwwbbbbrbbbbwwwrrr".

A palindrome is a word, phrase, or sequence that reads the same backwards as forwards. For example, "rggbwbggr", "b", "gg" are palindromes while "rgbb", "gbbgr" are not. Notice that an empty word, phrase, or sequence is palindrome.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct FASTINPUT {
  void f(string& s) {
    char buf[400005];
    long long a = scanf("%s", buf);
    a++;
    s = buf;
  }
  void f(long long& x) {
    long long a = scanf("%lld", &x);
    a++;
  }
  void f(int& x) {
    long long a = scanf("%d", &x);
    a++;
  }
  void f(double& x) {
    long long a = scanf("%lf", &x);
    a++;
  }
  template <typename A, typename B>
  void f(pair<A, B>& p) {
    f(p.first);
    f(p.second);
  }
  template <typename A>
  void f(vector<A>& x) {
    for (auto& y : x) f(y);
  }
  void read() {}
  template <typename Head, typename... Tail>
  void read(Head& H, Tail&... T) {
    f(H);
    read(T...);
  }
};
struct FASTOUTPUT {
  void f(string s) { printf("%s", s.c_str()); }
  void f(long long x) { printf("%lld", x); }
  void f(int x) { printf("%d", x); }
  void f(double x) { printf("%.20f", x); }
  void f(char x) { printf("%c", x); }
  void f(const char* a) { printf("%s", a); }
  template <typename A>
  void f(vector<A>& x) {
    for (auto& y : x) {
      f(y);
      f(" ");
    }
  }
  void print() {}
  template <typename Head, typename... Tail>
  void print(Head H, Tail... T) {
    f(H);
    print(T...);
  }
};
struct DEBUG {
  string open = "[", close = "]", sep = ", ";
  string f(string s) { return s; }
  string f(char c) { return string(1, c); }
  string f(int x) { return to_string(x); }
  string f(long long x) { return to_string(x); }
  string f(double x) { return to_string(x); }
  string f(const char* a) { return a; }
  template <typename A, typename B>
  string f(pair<A, B> p) {
    return open + f(p.first) + sep + f(p.second) + close;
  }
  template <typename A>
  string f(A v) {
    string s = open;
    for (auto a : v) {
      if (s != open) {
        s += sep;
      }
      s += f(a);
    }
    s += close;
    return s;
  }
  void show() { cout << endl; }
  template <typename Head, typename... Tail>
  void show(Head H, Tail... T) {
    cout << " " << f(H);
    show(T...);
  }
};
vector<long long> di = {0, 0, 1, -1, 1, -1, 1, -1};
vector<long long> dj = {1, -1, 0, 0, -1, -1, 1, 1};
long long ob(long long i, long long n) { return i < 0 || i >= n; }
long long tp(long long x) { return (1LL << x); }
long long roundup(long long a, long long b) {
  return a % b ? a / b + 1 : a / b;
}
template <typename A, typename B>
long long exist(A a, B b) {
  return a.find(b) != a.end();
}
vector<string> ssplit(string s) {
  vector<string> ans;
  stringstream ss;
  ss << s;
  while (ss >> s) ans.push_back(s);
  return ans;
}
void makemod(long long& x, long long m) {
  x %= m;
  x += m;
  x %= m;
}
long long powmod(long long a, long long b, long long m) {
  if (b == 0) return 1;
  long long h = powmod(a, b / 2, m);
  long long ans = h * h % m;
  return b % 2 ? ans * a % m : ans;
}
bool isvowel(char ch) {
  return (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');
}
long long sign(long long x) { return x > 0 ? 1 : x < 0 ? -1 : 0; }
void upmin(long long& x, long long v) { x = min(x, v); }
void upmax(long long& x, long long v) { x = max(x, v); }
const vector<long long> mods = {(long long)1e9 + 7, 998244353,
                                (long long)1e6 + 3, (long long)1e18 + 5};
const long long mod = mods[0];
const long long inf = mods[3];
const double eps = 1e-10;
const double pi = acos(0) * 2;
const string nl = "\n";
long long countodd(vector<long long>& A) {
  long long x = 0;
  for (auto a : A) {
    if (a < 0) return inf;
    if (a % 2) x++;
  }
  return x;
}
void solve() {
  vector<long long> A(4);
  do {
    FASTINPUT inp;
    inp.read(A);
  } while (false);
  if (countodd(A) <= 1) {
    do {
      FASTOUTPUT out;
      out.print("Yes", nl);
    } while (false);
    return;
  }
  A[0]--;
  A[1]--;
  A[2]--;
  A[3] += 3;
  if (countodd(A) <= 1) {
    do {
      FASTOUTPUT out;
      out.print("Yes", nl);
    } while (false);
  } else {
    do {
      FASTOUTPUT out;
      out.print("No", nl);
    } while (false);
  }
}
signed main() {
  long long t = 1;
  do {
    FASTINPUT inp;
    inp.read(t);
  } while (false);
  for (long long i = 0; i < t; i++) solve();
  return 0;
}